Question: Solve for $x$ : $ 3|x + 1| + 7 = 6|x + 1| + 6 $
Answer: Subtract $ {3|x + 1|} $ from both sides: $ \begin{eqnarray} 3|x + 1| + 7 &=& 6|x + 1| + 6 \\ \\ {- 3|x + 1|} && {- 3|x + 1|} \\ \\ 7 &=& 3|x + 1| + 6 \end{eqnarray} $ Subtract $6$ from both sides: $ \begin{eqnarray} 7 &=& 3|x + 1| + 6 \\ \\ {- 6} && {- 6} \\ \\ 1 &=& 3|x + 1| \end{eqnarray} $ Divide both sides by ${3}$ $ \dfrac{1} {{3}} = \dfrac{3|x + 1|} {{3}} $ Simplify: $ \dfrac{1}{3} = |x + 1| $ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ -\dfrac{1}{3} = x + 1 $ or $ \dfrac{1}{3} = x + 1 $ Solve for the solution where $x + 1$ is negative: $ - \dfrac{1}{3} = x + 1$ Subtract ${1}$ from both sides: $ \begin{eqnarray} - \dfrac{1}{3} &=& x + 1 \\ \\ {- 1} && {- 1} \\ \\ -\dfrac{1}{3} - 1 &=& x \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $3$ $ - \dfrac{1}{3} {- \dfrac{3}{3}} = x $ $ -\dfrac{4}{3} = x $ Then calculate the solution where $x + 1$ is positive: $ \dfrac{1}{3} = x + 1 $ Subtract ${1}$ from both sides: $ \begin{eqnarray} \dfrac{1}{3} &=& x + 1 \\ \\ {- 1} && {- 1} \\ \\ \dfrac{1}{3} - 1 &=& x \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $3$ $ \dfrac{1}{3} {- \dfrac{3}{3}} = x $ $ -\dfrac{2}{3} = x $ Thus, the correct answer is $x = -\dfrac{4}{3} $ or $x = -\dfrac{2}{3} $.